$\dfrac{ 2b + c }{ 3 } = \dfrac{ -4b - 4d }{ -5 }$ Solve for $b$.
Solution: Multiply both sides by the left denominator. $\dfrac{ 2b + c }{ {3} } = \dfrac{ -4b - 4d }{ -5 }$ ${3} \cdot \dfrac{ 2b + c }{ {3} } = {3} \cdot \dfrac{ -4b - 4d }{ -5 }$ $2b + c = {3} \cdot \dfrac { -4b - 4d }{ -5 }$ Multiply both sides by the right denominator. $2b + c = 3 \cdot \dfrac{ -4b - 4d }{ -{5} }$ $-{5} \cdot \left( 2b + c \right) = -{5} \cdot 3 \cdot \dfrac{ -4b - 4d }{ -{5} }$ $-{5} \cdot \left( 2b + c \right) = 3 \cdot \left( -4b - 4d \right)$ Distribute both sides $-{5} \cdot \left( 2b + c \right) = {3} \cdot \left( -4b - 4d \right)$ $-{10}b - {5}c = -{12}b - {12}d$ Combine $b$ terms on the left. $-{10b} - 5c = -{12b} - 12d$ ${2b} - 5c = -12d$ Move the $c$ term to the right. $2b - {5c} = -12d$ $2b = -12d + {5c}$ Isolate $b$ by dividing both sides by its coefficient. ${2}b = -12d + 5c$ $b = \dfrac{ -12d + 5c }{ {2} }$